An object is undergoing simple harmonic motion with frequency f = 3.5 Hz and an amplitude of 0.15 m. At t = 0.00 s the object is at x = 0.00 m. How long does it take the object to go from x = 0.00 m to x = 4.00×10−2 m.

Time taken by the object is 0.012 s .

Explanation:

Given :

Frequency , f = 3.5 Hz .

Amplitude , A = 0.15 m .

At time t = 0 , x = 0 m.

Since , at time t = 0 , x = 0 m .

Therefore , equation of displacement is :

    ...equation 1.

Here , is angular frequency and is given by :

We need to find the time at which its displacement is ,

Putting all these value in equation 1 we get ,

Hence , this is the required solution.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s